Disproof of the Reversion of the Fundamental Theorem of Cyclic Groups

Question

So the Fundamental Theorem of Cyclic Groups states that

Every subgroup of a cyclic group is cyclic.

In symbolic logic, we can state this as

$A: G\text{ is cyclic}$

$P: \forall S\subset G \text{ and }S\text{ is a group }, S \text{ is cyclic}$

Then $A\rightarrow P$

I was wondering, if the opposite is true, that is $P\rightarrow A$. I think that it's not true based on this counterexample:

Let $G=\lbrace a,b,c,d,e\rbrace $. We can draw a Cayley table to describe this group:

$$\begin{array}{c|ccc} \text{ } & a & b & c & d & e \\ \hline a & a& b & c & d &e \\ b & b & a & e & c &d \\ c & c & d & a & e &b \\ d & d & e & b & a & c \\ e & e & c & d & b & a \end{array}$$

All of the groups as far as I can see are cyclic, because they are of the form $\lbrace a, x\rbrace, x\in \lbrace b,c,d,e\rbrace$.

However, the whole group is obviously not cyclic because of the phenomenon described above ($x^2=a$).

Is this a valid counterexample to formally disprove the proposition?

Answer 1

As stated, this is trivially true, as $G$ is a subgroup of itself. So if every subgroup of $G$ is cyclic, $G$ itself must be cyclic.

If you're asking whether (every proper subgroup of $G$ is cyclic) implies $G$ cyclic, consider $G = \mathbb{Z}/2 \times \mathbb{Z}/2$. This group isn't cyclic, but all its proper subgroups have order $1$ or $2$, so they are cyclic.

Also, your example must be incorrect, since any group of order $5$ must be cyclic (as $5$ is prime).

Answer 2

For every prime $p$, $\mathbb{Z}_p \times \mathbb{Z}_p$ is not cyclic cause every element has order $p$ (or $1$). But every proper subgroup has order $p$ or $1$ because it must divide $p^2$, and thus is cyclic.

Answer 3

Your example is not a group: the operation is not associative (check $(ab)c$ and $a(bc)$, for example).

Moreover, any group of order 5 is cyclic, by Cauchy's theorem.

A simple counterexample is the Klein $4$-group, which is the non-cyclic group of order $4$: any proper subgroup must have order $<4$, and hence has prime order (or is trivial), and Cauchy's theorem again implies that all of these are cyclic.